3.1039 \(\int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=446 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (3 a^2 C+30 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{24 b d}+\frac {\sqrt {\sec (c+d x)} \left (a^2 (48 A+17 C)+42 a b B+8 b^2 (3 A+2 C)\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{24 d \sqrt {a+b \sec (c+d x)}}-\frac {\left (3 a^2 C+30 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{24 b d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\sqrt {\sec (c+d x)} \left (a^3 (-C)+6 a^2 b B+12 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{8 b d \sqrt {a+b \sec (c+d x)}}+\frac {(a C+2 b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{4 d}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{3 d} \]

[Out]

1/3*C*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/24*(42*a*b*B+8*b^2*(3*A+2*C)+a^2*(48*A+17*C))*(co
s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d
*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/d/(a+b*sec(d*x+c))^(1/2)+1/8*(6*a^2*b*B+8*b^3*B-a^3*C+12*a*b^2*(2*A+C))*(
cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*((b+a*
cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/b/d/(a+b*sec(d*x+c))^(1/2)+1/4*(2*B*b+C*a)*sec(d*x+c)^(3/2)*sin(d*x+
c)*(a+b*sec(d*x+c))^(1/2)/d-1/24*(24*A*b^2+30*B*a*b+3*C*a^2+16*C*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x
+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/b/d/((b+a*cos(d*x+c))/(a+
b))^(1/2)/sec(d*x+c)^(1/2)+1/24*(24*A*b^2+30*B*a*b+3*C*a^2+16*C*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)*(a+b*sec(d*x+
c))^(1/2)/b/d

________________________________________________________________________________________

Rubi [A]  time = 1.65, antiderivative size = 446, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {4096, 4102, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (3 a^2 C+30 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{24 b d}+\frac {\sqrt {\sec (c+d x)} \left (a^2 (48 A+17 C)+42 a b B+8 b^2 (3 A+2 C)\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{24 d \sqrt {a+b \sec (c+d x)}}-\frac {\left (3 a^2 C+30 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{24 b d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\sqrt {\sec (c+d x)} \left (6 a^2 b B+a^3 (-C)+12 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{8 b d \sqrt {a+b \sec (c+d x)}}+\frac {(a C+2 b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{4 d}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((42*a*b*B + 8*b^2*(3*A + 2*C) + a^2*(48*A + 17*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2,
(2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(24*d*Sqrt[a + b*Sec[c + d*x]]) + ((6*a^2*b*B + 8*b^3*B - a^3*C + 12*a*b^2*
(2*A + C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(8
*b*d*Sqrt[a + b*Sec[c + d*x]]) - ((24*A*b^2 + 30*a*b*B + 3*a^2*C + 16*b^2*C)*EllipticE[(c + d*x)/2, (2*a)/(a +
b)]*Sqrt[a + b*Sec[c + d*x]])/(24*b*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + ((24*A*b^2 + 3
0*a*b*B + 3*a^2*C + 16*b^2*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(24*b*d) + ((2*b*B + a
*C)*Sec[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(4*d) + (C*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x
])^(3/2)*Sin[c + d*x])/(3*d)

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 4096

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(m + n + 1), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^
n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a*C
*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] &&
!LeQ[n, -1]

Rule 4102

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 0]

Rule 4108

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} \int \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \left (\frac {1}{2} a (6 A+C)+(3 A b+3 a B+2 b C) \sec (c+d x)+\frac {3}{2} (2 b B+a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {(2 b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{6} \int \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{4} a (24 a A+6 b B+7 a C)+\frac {1}{2} \left (12 a^2 B+6 b^2 B+a b (24 A+13 C)\right ) \sec (c+d x)+\frac {1}{4} \left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {\left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right ) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 b d}+\frac {(2 b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {\int \frac {-\frac {1}{8} a \left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right )+\frac {1}{4} a b (24 a A+6 b B+7 a C) \sec (c+d x)+\frac {3}{8} \left (6 a^2 b B+8 b^3 B-a^3 C+12 a b^2 (2 A+C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{6 b}\\ &=\frac {\left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right ) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 b d}+\frac {(2 b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {\int \frac {-\frac {1}{8} a \left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right )+\frac {1}{4} a b (24 a A+6 b B+7 a C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{6 b}+\frac {\left (6 a^2 b B+8 b^3 B-a^3 C+12 a b^2 (2 A+C)\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{16 b}\\ &=\frac {\left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right ) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 b d}+\frac {(2 b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {\left (-24 A b^2-30 a b B-3 a^2 C-16 b^2 C\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{48 b}+\frac {1}{48} \left (42 a b B+8 b^2 (3 A+2 C)+a^2 (48 A+17 C)\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {\left (\left (6 a^2 b B+8 b^3 B-a^3 C+12 a b^2 (2 A+C)\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{16 b \sqrt {a+b \sec (c+d x)}}\\ &=\frac {\left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right ) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 b d}+\frac {(2 b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {\left (\left (42 a b B+8 b^2 (3 A+2 C)+a^2 (48 A+17 C)\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{48 \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (6 a^2 b B+8 b^3 B-a^3 C+12 a b^2 (2 A+C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{16 b \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (-24 A b^2-30 a b B-3 a^2 C-16 b^2 C\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{48 b \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {\left (6 a^2 b B+8 b^3 B-a^3 C+12 a b^2 (2 A+C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{8 b d \sqrt {a+b \sec (c+d x)}}+\frac {\left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right ) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 b d}+\frac {(2 b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {\left (\left (42 a b B+8 b^2 (3 A+2 C)+a^2 (48 A+17 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{48 \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (-24 A b^2-30 a b B-3 a^2 C-16 b^2 C\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{48 b \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=\frac {\left (42 a b B+8 b^2 (3 A+2 C)+a^2 (48 A+17 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{24 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (6 a^2 b B+8 b^3 B-a^3 C+12 a b^2 (2 A+C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{8 b d \sqrt {a+b \sec (c+d x)}}-\frac {\left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{24 b d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {\left (24 A b^2+30 a b B+3 a^2 C+16 b^2 C\right ) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 b d}+\frac {(2 b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [C]  time = 6.98, size = 800, normalized size = 1.79 \[ \frac {(a+b \sec (c+d x))^{3/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac {1}{6} (6 b B \sin (c+d x)+7 a C \sin (c+d x)) \sec ^2(c+d x)+\frac {2}{3} b C \tan (c+d x) \sec ^2(c+d x)+\frac {\left (3 C \sin (c+d x) a^2+30 b B \sin (c+d x) a+24 A b^2 \sin (c+d x)+16 b^2 C \sin (c+d x)\right ) \sec (c+d x)}{12 b}\right )}{d (b+a \cos (c+d x)) (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}-\frac {(a+b \sec (c+d x))^{3/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac {2 \left (-96 A b a^2-28 b C a^2-24 b^2 B a\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{\sqrt {b+a \cos (c+d x)}}+\frac {2 \left (9 C a^3-6 b B a^2-120 A b^2 a-56 b^2 C a-48 b^3 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{\sqrt {b+a \cos (c+d x)}}+\frac {2 i \left (3 C a^3+30 b B a^2+24 A b^2 a+16 b^2 C a\right ) \sqrt {\frac {a-a \cos (c+d x)}{a+b}} \sqrt {\frac {\cos (c+d x) a+a}{a-b}} \cos (2 (c+d x)) \left (a \left (2 b F\left (i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {b-a}{a+b}\right )+a \Pi \left (1-\frac {a}{b};i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {b-a}{a+b}\right )\right )-2 b (a+b) E\left (i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {b-a}{a+b}\right )\right ) \sin (c+d x)}{\sqrt {\frac {1}{a-b}} b \sqrt {1-\cos ^2(c+d x)} \sqrt {\frac {a^2-a^2 \cos ^2(c+d x)}{a^2}} \left (-a^2+2 b^2+2 (b+a \cos (c+d x))^2-4 b (b+a \cos (c+d x))\right )}\right )}{48 b d (b+a \cos (c+d x))^{3/2} (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

-1/48*((a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(-96*a^2*A*b - 24*a*b^2*B - 28*a
^2*b*C)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]] + (
2*(-120*a*A*b^2 - 6*a^2*b*B - 48*b^3*B + 9*a^3*C - 56*a*b^2*C)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2
, (c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]] + ((2*I)*(24*a*A*b^2 + 30*a^2*b*B + 3*a^3*C + 16*a*b^2
*C)*Sqrt[(a - a*Cos[c + d*x])/(a + b)]*Sqrt[(a + a*Cos[c + d*x])/(a - b)]*Cos[2*(c + d*x)]*(-2*b*(a + b)*Ellip
ticE[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*(2*b*EllipticF[I*ArcSinh[Sq
rt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*EllipticPi[1 - a/b, I*ArcSinh[Sqrt[(a - b)^(
-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)]))*Sin[c + d*x])/(Sqrt[(a - b)^(-1)]*b*Sqrt[1 - Cos[c + d*x]^
2]*Sqrt[(a^2 - a^2*Cos[c + d*x]^2)/a^2]*(-a^2 + 2*b^2 - 4*b*(b + a*Cos[c + d*x]) + 2*(b + a*Cos[c + d*x])^2)))
)/(b*d*(b + a*Cos[c + d*x])^(3/2)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + ((a
+ b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((Sec[c + d*x]^2*(6*b*B*Sin[c + d*x] + 7*a*C*S
in[c + d*x]))/6 + (Sec[c + d*x]*(24*A*b^2*Sin[c + d*x] + 30*a*b*B*Sin[c + d*x] + 3*a^2*C*Sin[c + d*x] + 16*b^2
*C*Sin[c + d*x]))/(12*b) + (2*b*C*Sec[c + d*x]^2*Tan[c + d*x])/3))/(d*(b + a*Cos[c + d*x])*(A + 2*C + 2*B*Cos[
c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/2)*sqrt(sec(d*x + c)), x)

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maple [C]  time = 2.38, size = 5245, normalized size = 11.76 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/2)*sqrt(sec(d*x + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int((a + b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*sec(d*x+c)**(1/2),x)

[Out]

Timed out

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